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NEET UG Ā· Physics

Nuclei

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NEET UG Ā· Physics

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Check binding-mass conventions, half-life fractions and total nuclear energy balances. Nuclei: count particles and keep the energy basis clear

About 7 min Ā· NEET UG

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Nuclei: count particles and keep the energy basis clear

Check binding-mass conventions, half-life fractions and total nuclear energy balances.

What you will understand

For NEET UG Physics students: count protons and neutrons, use a consistent mass convention for binding energy, separate remaining nuclei from decayed nuclei and calculate nuclear energy changes from total binding.

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Quick answer: conserve nucleon count and keep the mass reference consistent

Mass number A counts protons plus neutrons; atomic number Z counts protons. A bound nucleus has lower rest mass than its separated constituent nucleons, and its positive binding energy is the energy required to separate them. Use either nuclear masses with bare protons or a consistent atomic-mass formula with hydrogen atoms. Radioactive half-life concerns a remaining-parent population, while activity also includes the decay constant. Total binding and binding per nucleon answer different questions.

Helium binding: an atomic-mass convention cancels electron rest masses

Original helium-4 binding-energy bookkeeping illustration. Supplied rounded masses are neutral hydrogen atom 1.007825 u, neutron 1.008665 u and neutral helium-4 atom 4.002603 u. Two hydrogen atom masses plus two neutron masses total 4.032980 u; subtracting helium atom mass gives 0.030377 u. Multiplying by supplied 931.5 MeV per u c squared gives approximately 28.30 MeV total binding energy, or 7.07 MeV per nucleon. This atomic-mass convention cancels two electron rest masses; small electronic binding-energy corrections are neglected at this precision. Do not subtract an atomic mass from a sum of bare proton masses without correcting the electrons. The difference reflects the mass-energy of the bound system, not missing protons or neutrons.
The supplied hydrogen and helium masses are neutral-atom masses. Two hydrogen atoms supply the same number of electron rest masses as one helium atom, so these cancel in this approximation. The resulting positive mass difference gives about 28.30 MeV total binding, or 7.07 MeV per nucleon. Small electronic binding corrections are neglected.

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Compute total binding before dividing by the nucleon count

Use supplied rounded masses mH=1.007825 u for a neutral hydrogen atom, mn=1.008665 u and mHe=4.002603 u for a neutral helium-4 atom. With Z=2 and A=4,

Ī”m=2mH+2mnāˆ’mHe=0.030377 u\Delta m=2m_H+2m_n-m_\mathrm{He}=0.030377\,\mathrm{u}
. With supplied 1 u c2=931.5 MeV1\,\mathrm{u}\,c^2=931.5\,\mathrm{MeV}, total binding is about 28.296 MeV, or 28.30 MeV to the displayed precision. Binding per nucleon is about 7.07 MeV. The atomic-mass expression is Bā‰ˆ[ZmH+(Aāˆ’Z)mnāˆ’Matom]c2B\approx[Zm_H+(A-Z)m_n-M_\mathrm{atom}]c^2. Electron rest masses cancel; electronic binding corrections are neglected at this precision. Do not replace mH with a bare-proton mass while retaining the neutral-atom mass on the other side.

A, Z and radius describe different quantities

For a nuclide with A=56 and Z=26, proton count is 26 and neutron count is 30. Isotopes share Z but differ in neutron count; isobars share A, and isotones share neutron count. In the approximate radius law R=R0A1/3R=R_0A^{1/3}, increasing A by a factor of eight doubles R rather than multiplying it by eight. This scaling makes volume approximately proportional to nucleon count and density roughly independent of A in the simple model. A mass number is a count, not an exact atomic mass in u.

Half-life describes the remaining parent population

Original plot of expected remaining-parent fraction N/N0=2 to the power minus elapsed half-lives, from zero to three half-lives. The graph has a common zero baseline; fractions 1, one half, one quarter and one eighth occur at zero, one, two and three half-lives. In the supplied example T=2 hours and initial expected count N0=8000, six hours leaves an expectation of 1000 parent nuclei and 7000 parent decays. Equal intervals remove half of what remains, not an equal absolute number. This is a constant-rate single-parent ensemble model, with no replenishment; finite samples fluctuate and daughter activity is not included. A particular nucleus does not have a predetermined decay appointment.
The graph plots the expected remaining fraction against elapsed half-lives. For a single parent with T=2 h, the illustrative expectation 8000 becomes 1000 after 6 h; 7000 parents have decayed. Equal intervals halve what remains. The smooth curve describes a large-population model; actual finite counts fluctuate, and daughter activity is not included.

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Remaining fraction, decayed fraction and activity

Suppose a single parent species has constant half-life T=2.00 h, initial expected population N0=8000 and no replenishment. After 6.00 h, three half-lives have elapsed, so N/N0=2āˆ’3=1/8N/N_0=2^{-3}=1/8. The expected remaining count is 1000, and the expected number of parent decays is 7000. The decayed fraction is 7/8, not 1/8. Since Ī»=ln⁔2/T\lambda=\ln2/T, the initial activity is about 0.770 Bq and after 6 h about 0.0963 Bq, using T=7200 s and A=Ī»N\mathcal A=\lambda N. Both count and parent activity fall by the same fraction for this one species. A Bq is one decay per second; this does not specify a dose or the combined activity of radioactive daughters.

Decay bookkeeping does not by itself prove that a decay occurs

An alpha emission changes A to Aāˆ’4 and Z to Zāˆ’2. A beta-minus emission leaves A unchanged and raises Z by one, with an electron and an electron antineutrino in the final state. An ordinary gamma transition changes nuclear excitation without changing A or Z. For a hypothetical parent with A=240,Z=90, one alpha followed by one beta-minus gives A=236,Z=89. That is nucleon/charge bookkeeping; energetic feasibility and actual branching require more information. The emitted beta electron is not evidence of a pre-existing electron stored inside the nucleus.

Use total binding to find a supplied reaction energy

Consider a supplied hypothetical balanced fission-like comparison with one A=100 nucleus becoming two A=50 fragments, conserved proton/neutron totals and no additional particles. Parent binding per nucleon is 8.00 MeV and each fragment value is 8.50 MeV. Parent total binding is 800 MeV; final total is 2(50)(8.50)=850 MeV2(50)(8.50)=850\,\mathrm{MeV}. Thus the rest-mass energy decrease is

Q=Bfinalāˆ’Binitial=+50 MeVQ=B_\mathrm{final}-B_\mathrm{initial}=+50\,\mathrm{MeV}
. The positive value is available as product kinetic energy or radiation according to the stated final states. It does not prove that a real isotope exists with these values, that the reaction proceeds spontaneously or what its rate is. Using only 8.50āˆ’8.00=0.50 MeV loses the nucleon weighting.

One half-life does not schedule each nucleus

The exponential model assigns a decay probability per unit time to each remaining nucleus. It does not specify which nucleus will decay next or say every nucleus survives exactly one half-life. A nucleus that remains after one half-life is still described by the same constant decay probability in this model. The remaining population falls smoothly in the large-sample approximation, while individual events are random.

Common mistakes: total, per nucleon, remaining and activity are different

Do not mix atomic and nuclear masses without correcting the electron convention. Positive binding energy does not mean the bound system is heavier. Multiply binding per nucleon by the relevant nucleon count before comparing totals. After three half-lives, one eighth remains and seven eighths has decayed. Activity depends on both decay constant and parent count. A balanced A/Z equation is necessary bookkeeping, not proof of an energetically allowed channel.

Try it: decayed is not remaining

A single parent species has half-life 2 h. After 4 h, what fractions remain and have decayed in the ensemble model? Does the remaining fraction imply that every original nucleus had a fixed four-hour lifetime?

Reveal the answer

Two half-lives leave 1/4 of the expected initial parent population; 3/4 has decayed. This is a population/probability statement. It does not assign a fixed lifetime to each original nucleus.

Try it: alpha then beta-minus

For a hypothetical parent with A=210 and Z=84, perform A/Z bookkeeping for one alpha emission followed by one beta-minus emission. What are the final A, Z and neutron count? Does bookkeeping alone establish that the chain is physically allowed?

Reveal the answer

After alpha emission A=206,Z=82. Beta-minus leaves A=206 and raises Z to 83, so neutron count is 206āˆ’83=123. Electron and antineutrino must be included in the beta-minus process. This is a bookkeeping exercise; actual energy differences and available decay branches are needed to establish a physical chain.

Come back tomorrow: name the mass, count and energy basis

Write a binding-energy balance with one consistent mass convention. Then distinguish remaining population, decayed population and activity without looking at the graph. Revisit NEET Atoms to compare electronic and nuclear energy changes, or NEET Work, Energy and Power for initial/final energy bookkeeping.

Chapter practice opens the broad Nuclei pool.

Sources and review

Technical review: Codex (automated). Primary textbook sources checked on 23 October 2026. Diagrams, worked examples and self-checks are original. Numerical calculations use stated approximations and supplied rounded constants. Human subject review is not claimed. Chapter catalogues and these explanations do not establish complete coverage of every current examination format or syllabus.

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